Potassium ions (charge -e, mass mK = 39 mp) are accelerated from rest at plate A

Question

Potassium ions (charge -e, mass mK = 39 mp) are accelerated from rest at plate A through a voltage difference of ?V = 5 kV to plate B. Some of the ions pass through a small slit in plate B to enter into a region of uniform magnetic field (the shaded region in the diagram). mp is the mass of a proton, 1.67 × 10–27 kg. a) Which plate is at a higher potential, A or B? Calculate the speed of the K- ions when they reach the slit in plate B. b) When the ions enter the region of constant magnetic field, they experience a force that is perpendicular to their velocity. Using Newton’s Laws, what must be the magnitude and direction of this force if these ions are to follow a curved path of radius r = 10 cm? c) Find the magnitude and direction of the magnetic field needed for the force you found in part (b).

2. Putting It Together: The Mass Spectrometer (3 pts) Potassium ions (charge -e, mass mk- 39 mp) are accelerated from rest at plate A through a voltage difference of AV-5 kV to plate B. Some of the ions pass through a small slit in plate B to enter into a region of uniform magnetic field (the shaded region in the diagram). mp is the mass of a proton, 1.67 x 1027 kg AV 0 a) Which plate is at a higher potential, A or B? Calculate the speed of the K ions when they reach the slit in plate B Detector b) When the ions enter the region of constant magnetic field, 10 cm they experience a force that is perpendicular to their velocity. Using Newton's Laws, what must be the magnitude and direction of this force if these ions are to follow a curved path of radius r 10 cm? c) Find the magnitude and direction of the magnetic field needed for the force you found in part (b)

Explanation / Answer

a) Negative charge gets attracted by higher potential, hence plate B has higher potential.

KE=qV

1/2mv^2= qV

1/2(39*1.67*10^-27)v^2 = (1.6*10^-19)(5000) => v= 1.57*10^5 m/s

b) Direction of B must be into the page thus by Lorentz force(magnetic force) acts in the downward direction..

Applying Newton’s 2nd law we get,

FB =mv^2/r = (39*1.67*10^-27*(1.57*10^5)^2)/0.1 =

FB = 1.6*10^-14 N

c)

FB = qvB

1.6*10^-14 =(1.6*10^-19)( 1.57*10^5)(B)     => B= 0.637 T

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